Answer:
When the Earth and sun are perfectly lined up, then it will happen. They can tell when it's going to happen.
Explanation:
This is why it only happens in some places. Some days it's not sunny out, so it's not going to happen.
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>
![y_{f}=0m](https://tex.z-dn.net/?f=y_%7Bf%7D%3D0m)
![v_{iD}=0m/s](https://tex.z-dn.net/?f=v_%7BiD%7D%3D0m%2Fs)
, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so ![t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s](https://tex.z-dn.net/?f=t_%7BE%7D%3Dt_%7BD%7D-t_%7BT%7D%3D7.82s-5.4s%3D2.42s)
Answer:
5,000J
Explanation:
Work = Force x Distance
Distance back and forth is canceled out, so either the answer is + or -
5.0m + 5.0m = 10.0m
500N x 10.0m = 5,000J
Answer:
A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction
Explanation:
The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³